Let $f(x)=\sin(x)x^{-2}$. Find $f'(x)$. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{\sin(x)}{x^2}-\dfrac{2\cos(x)}{x}$ (Choice B) B $\dfrac{\sin(x)}{x^2}-\dfrac{2\cos(x)}{x^3}$ (Choice C) C $\dfrac{\cos(x)}{x^2}-\dfrac{2\sin(x)}{x}$ (Choice D) D $\dfrac{\cos(x)}{x^2}-\dfrac{2\sin(x)}{x^3}$
$f(x)$ is the product of two, more basic, expressions: $\sin(x)$ and $x^{-2}$. Therefore, the derivative of $f$ can be found using the product rule : $\begin{aligned} \dfrac{d}{dx}[u(x)v(x)]&=\dfrac{d}{dx}[u(x)]v(x)+u(x)\dfrac{d}{dx}[v(x)] \\\\ &=u'(x)v(x)+u(x)v'(x) \end{aligned}$ $\begin{aligned} &\phantom{=}f'(x) \\\\ &=\dfrac{d}{dx}\left[\sin(x)x^{-2}\right] \\\\ &=\dfrac{d}{dx}\left[\sin(x)\right]x^{-2}+\sin(x)\dfrac{d}{dx}[x^{-2}]&&\gray{\text{The product rule}} \\\\ &=\cos(x)x^{-2}+\sin(x)(-2)x^{-3}&&\gray{\text{Differentiate }\sin(x)\text{ and }x^{-2}} \\\\ &=\dfrac{\cos(x)}{x^2}-\dfrac{2\sin(x)}{x^3}&&\gray{\text{Simplify}} \end{aligned}$ In conclusion, $f'(x)=\dfrac{\cos(x)}{x^2}-\dfrac{2\sin(x)}{x^3}$